Writing \(\displaystyle{{\tan}^{{2}}{\left({2}{{\sec}^{{-{1}}}{\left({\frac{{{x}}}{{{3}}}}\right)}}\right)}}\) in algebraic form

talpajocotefnf3

talpajocotefnf3

Answered question

2022-03-23

Writing tan2(2sec1(x3)) in algebraic form

Answer & Explanation

Nunnaxf18

Nunnaxf18

Beginner2022-03-24Added 18 answers

Let θ=sec1(x3) thus secθ=x3  cosθ=3x since sec=1cos
You have cosθ=3x. So sinθ=19x2 and tanθ=19x23x=x293
So tan2(2θ)=(2tanθ1tan2θ)2=
4x299(1x299)2=

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